Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 - - PowerPoint PPT Presentation

stopping power of 0 5 1 5mev protons in mylar
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Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 - - PowerPoint PPT Presentation

Stopping Power of 0.5-1.5MeV Protons in Mylar Tinyiko S. Maluleke 1 Supervisors: C. Pineda-Vargas 2 M. Msimanga 2 V. Kuzmin 3 1 Stellenbosch University, 2 iThemba Labs, 3 JINR AIM To measure the experimental stopping power for protons


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SLIDE 1

Stopping Power of 0.5-1.5MeV Protons in Mylar

Tinyiko S. Maluleke1 Supervisors: C. Pineda-Vargas2

  • M. Msimanga2
  • V. Kuzmin3

1 Stellenbosch University, 2 iThemba Labs, 3 JINR

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SLIDE 2

20090924 South African Summer Student Practice '09 2

AIM

  • To measure the experimental stopping power

for protons in mylar (C10H8O4).

  • To compare the current results with theoretical

calculations and previous experimental results.

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SLIDE 3

20090924 South African Summer Student Practice '09 3

INTRODUCTION

  • Stopping Power S:
  • It can be expressed as sum of electronic and

nuclear stopping power:

S= 1 N dE dx S=SeSn

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SLIDE 4

20090924 South African Summer Student Practice '09 4

EXPERIMENTAL SETUP

  • Beam: 0.5-

1.5MeV

  • Foil Thickness
  • Mylar: 1.68μm
  • Primary Target:

Pt/C

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SLIDE 5

20090924 South African Summer Student Practice '09 5

RESULTS (RBS Spectrum)

  • C1 – Pt-scattered 1H+
  • C2 – Pt-scattered 1H+

through the foil.

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SLIDE 6

20090924 South African Summer Student Practice '09 6

RESULTS (Calibration)

  • Scattered proton energy:

where kinematic factor K: θ – Scattering angle M1 – Projectile mass M2 – Target nuclei mass

E1=KE0

K=[

M 2

2−M 1 2sin 2 1 2M 2cos

M 1M 2

]

2

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SLIDE 7

20090924 South African Summer Student Practice '09 7

RESULTS (Stopping Power)

  • The energy loss
  • f protons in

mylar foil is rather small (less than 20% of E1).

  • Stopping power is

evaluated at average energy, Eavg.

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SLIDE 8

20090924 South African Summer Student Practice '09 8

RESULTS

  • Stopping power:

if where: and is foil thickness, at

S≈ E x

 E=E1−E2

Eavg= E1E 2 2  E E1 0.20

 x

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SLIDE 9

20090924 South African Summer Student Practice '09 9

CONCLUSION

  • Current results deviates from theory and

previous results by 5-10%.

  • There may be error in thickness of foil.

SIMNRA calculations gives mass of foil as 1.82μm, which yield good correspondence.

  • Therefore mass of foil has to be comfirmed by

weighing method.

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SLIDE 10

20090924 South African Summer Student Practice '09 10

THANK YOU!